# 5×5 sudoku with special properties

Complete the network with numbers from 1 to 5 so that the same number does not occur in any line, column, specially marked figure and on both diagonals.

I tried to start with a number that I know for sure goes to that place, but I couldn't find one. The only thing I've tried is guessing the numbers on the diagonal, but there's probably a much better solution that I'm still not seeing.

Source: School Sudoku competition for children from 1st to 4th grade of primary school in Croatia

• For similar puzzles, please visit: puzzling.stackexchange.com/questions/117064/… Oct 15, 2022 at 10:03
• The funny thing is that if you completely ignore the regions, there are sill only 2 solutions. If you ignore the digits, all solutions are permutations of the 2 solutions, i.e. solutions where digits are replaced by others. Oct 15, 2022 at 13:47
• Need the full rules. I presume it is that each row, column, and region must be 1 to 5. But do the marked diagonals also need to be 1 to 5? Oct 16, 2022 at 14:22
• I make videos on Math puzzles and post them on YouTube . I want to make a video on this puzzle . Is this an open source puzzle or would I need permission from School sudoku competition to make a video on this that goes on YouTube ? Jan 5, 2023 at 6:35

Remember that each shape has 5 squares, so must contain all values 1 to 5 exactly once.

There is only one place where you can

put a 5 in the figure in the middle left. The 5 cannot go in R3C2 or R4C3 so must go in the only other empty spot of this figure, R3C1.

Simmilarly,

The figure at the top right has only one place where its 3 can go.

• Also R4C3 having to be a 1(naked single) Oct 15, 2022 at 8:45
• Why does a partial solution get chosen as the answer? Oct 16, 2022 at 13:58
• @BillOnne If you read the question, it appears OP was not posting this as a "contest" puzzle, but genuinely wanted help solving. Oct 18, 2022 at 16:14

As Jaap stated,

the middle-left region's 5 must be R3C1

and

the top right region's 3 must be R2C5

This allows us to complete row 2 as

42513.

the middle-left region as

R4C3=1, so R3C2=4

so row 3 must be

54321

so row 4 must be

35142

so column 2 must be

32451 as otherwise the bottom left region has two 3s

and column 3 must be

25314 as otherwise the top left region has two 4s

and column 4 must be

51243 as otherwise the top left region has two 3s

after which the remaining cells can be determined as each region must have 1, 2, 3, 4, 5 once each.

We don't need the diagonal constraint. Without it, we can determine row 3 and R4C3. The top right region's

3 must be in C5, so R4C5 is not 3 but 2, so row 4 is 35142.

This determines column 2 as above, locating the final

1 at R1C1. So R2C1 is not 2 but 4, so column 1 is 14532, so R5C3 is 4 to complete the bottom left region.

This determines column 3; this time the reason is that

52314 would give row 2 two 2s.

And the rest is easy.

Took me less than 5 minutes using pen and paper.

13254
42513
54321
35142
21435

(Sorry, I don’t know how to make it look like a graph)

• Why does a full answer get down-voted without comment? Oct 16, 2022 at 13:59
• @BillOnne Possibly because the answerer showed no working, so the answer gives no clue as to how one was supposed to find it. Oct 19, 2022 at 6:54
• @RosieF: Thank you for the feedback. I’ll keep that in mind for my future answers! :) Oct 20, 2022 at 0:04